没有 numpy 的列表切片

Question

I have the ohlc list as below:

ohlc = [["open", "high", "low", "close"],
        [100, 110, 70, 100],
        [200, 210, 180, 190],
        [300, 310, 300, 310]]

I want to slice it as:

[["open"],[100],[200],[300]]

We can easily slice that list using numpy , but I don't know how to do it without numpy 's help.

I tried the method listed below but it didn't show the value I wanted:

ohlc[:][0]
ohlc[:][:1]
ohlc[0][:]

Answer 1

The zip function gets you tuples containing elements from the i -th index of every sublist:

In [217]: ohlc = [["open", "high", "low", "close"], 
     ...:         [100, 110, 70, 100], 
     ...:         [200, 210, 180, 190], 
     ...:         [300, 310, 300, 310]] 
     ...:

In [218]: for t in zip(*ohlc): print(t)
('open', 100, 200, 300)
('high', 110, 210, 310)
('low', 70, 180, 300)
('close', 100, 190, 310)

You're looking for the first one of these, you call on your friend next() .

In [219]: next(zip(*ohlc))                                                                                                                                                                                                 
Out[219]: ('open', 100, 200, 300)

But that's just a single tuple with all the elements and not a list of lists like you wanted, so use a list comprehension:

In [220]: [[t] for t in next(zip(*ohlc))]
Out[220]: [['open'], [100], [200], [300]]

Answer 2

You can iterate over the list and take the element in index in every sub list

ohlc = [["open", "high", "low", "close"],
        [100, 110, 70, 100],
        [200, 210, 180, 190],
        [300, 310, 300, 310]]

index = 0
result = [[o[index]] for o in ohlc] # [['open'], [100], [200], [300]]

Answer 3

Can it be the right solution this?

list_ = []
for i in ohlc:
   list_.append((i[0]))

Answer 4

def slice_list(input):
    ans = []
    for x in input:
        ans.append(x[0])
    return ans